Common Exam - 2006

Department of PhysicsUniversity of UtahAugust 26, 2006

Examination booklets have been provided for recording your work and your solutions.

Please note that there is a separate booklet for each numbered question (i.e., usebooklet #1 for problem #1, etc.).

To receive full credit, not only should the correct solutions be given, but a sufficient

number of steps should be given so that a faculty grader can follow your reasoning.

Define all algebraic symbols that you introduce. If you are short of time it may be helpful

to give a clear outline of the steps you intended to complete to reach a solution. In some

of the questions with multiple parts you will need the answer to an earlier part in order to

work a later part. If you fail to solve the earlier part you may represent its answer with an

algebraic symbol and proceed to give an algebraic answer to the later part. This is a

closed book exam: No notes, books, or other records should be consulted. YOU MAY

ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided

equally among the ten questions of the examination.

All work done on scratch paper should be NEATLY transferred to answer booklets.

SESSION 1

COMMON EXAM DATA SHEET

e = - 1.60 × 10 C = - 4.80 × 10 esu-19 -10

c = 3.00 × 10 m/s = 3.00 × 10 cm/s8 10

h = 6.64 × 10 JAs = 6.64 × 10 ergAs = 4.14 × 10 MeVAs -34 -27 -21

S = 1.06 × 10 JAs = 1.06 × 10 ergAs = 6.59 × 10 MeVAs -34 -27 -22

k = 1.38 × 10 J/K = 1.38 × 10 erg/K -23 -16

g = 9.80 m/s = 980 cm/s2 2

G = 6.67 × 10 NAm /kg = 6.67 × 10 dyneAcm /g-11 2 2 -8 2 2

AN = 6.02 × 10 particles/gmAmole = 6.02 × 10 particles/kgAmole23 26

og (SI units) = 8.85 × 10 F/m -12

o: (SI units) = 4B × 10 H/m -7

m(electron) = 9.11 × 10 kg = 9.11 × 10 g= 5.4859 × 10 AMU = 511 keV -31 -28 -4

M(proton) 1.673 × 10 kg = 1.673 × 10 g = 1.0072766 AMU = 938.2 MeV -27 -24

M(neutron) 1.675 × 10 kg = 1.675 × 10 g = 1.0086652 AMU = 939.5 MeV -27 -24

M(muon) = 1.88 × 10 kg = 1.88 × 10 g -28 -25

1 mile = 1609 m

1 m = 3.28 ft

1 eV = 1.6 × 10 J = 1.6 × 10 ergs -19 -12

hc = 12,400 eVAD

Table of Integrals and Other Formulas

Spherical Harmonics

Conic Section

(origin at the focus; e = 1 for parabola)

Normal Distribution

Cylindrical Coordinates (orthonormal bases)

Spherical Coordinates (orthonormal bases)

Maxwell’s Equations (Rationalized MKS)

Maxwell’s Equations (Gaussian Units)

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Problem 3 - Mechanics:a) [4 points] Express the moment of inertia with respect to the yo-yo rotation axis.

The moment of inertia is obtained by summing all the mass elements multiplied by their squared distance to the 

considered axis. Here, taking advantage of the symmetry we get  I= MR2∫

0

R

r2⋅2 r⋅dr= M⋅R2

2b) [4 points] Express the yo-yo total energy in terms of its speed v , distance z

from the starting point, mass M , radius R , core radius r and gravitation acceleration g (make sure to include the rotation kinetic, the translation kinetic energy and the gravitational potential energy).The total energy  E consists of three terms: the translation kinetic energy  K T , the rotation kinetic energy  K R and the gravitation potential energy U g . The translation kinetic energy and the gravitation 

potential energy simply are  K T=12

M v2

and  U g=−M⋅g⋅z . The rotation kinetic energy is 

K R=12

I⋅2where  =

vr

is the angular velocity and  I is the moment of inertia.  Using the 

result from question a) we obtain  K R=M⋅R2⋅v2

4r 2 . 

Putting all the contributions together, the total energy is  E=12

M v2M⋅R2⋅v2

4r 2 −M⋅g⋅z

c) [4 points] Establish the equation of motion z t for the yo-yo.

The total energy must be conserved and dEdt

=0 . With  v= z=dzdt

and  z=d 2 z

dt 2 , we have 

0=M z zM⋅R2

2r2 z z−M⋅g⋅z from which, solving for  z , we obtain  z=2g⋅r2

2r2R2 . 

The motion is uniformly accelerated and  z t =12

z⋅t 2. 

d) [4 points] Establish tension of the string when the yo-yo is going down?The yo­yo is subject to its weight  M⋅g and to the tension of the string  T . Applying the fundamental 

relation of dynamics, we write  T −M⋅g=−M z and  T =−M⋅g⋅R2

2 r 2R2=−M⋅g

12r2 /R2 .  

e) [5 points] After moving down by a distance l , the yo-yo reaches the end of the string and starts moving up. Estimate the average acceleration of the yo-yo during the time it takes for the motion to change direction (half a turn).The motion is uniformly accelerated, so after going down a distance  l , the downward speed of the yo­yo is 

v l=2 l z . The corresponding angular velocity is  l=v l

r. The yo­yo goes half a turn in a time 

t 1/2=l

and then starts going up with the speed  v l . In the time the motion changes direction, the 

acceleration is of the order of  a=2v l

t1 /2=4l z

r. 

f) [4 points] The yo-yo moves up and down in cycles. Make a qualitative graph of

z t ,dzdt

t andd 2 z

dt 2 t for at least one cycle.

On the graph, I represent one cycle. At first,  z t increases quadratically with time until the end of the string is reached. The the motion changes direction and  z t

decreases quadratically to 0. Correspondingly, dzdt

t

always increases linearly with time and  change sign when 

the end of the string is reached. d 2 z

dt 2 t is constant 

except at the end of the string. dz/dt

t

t

d2z/dt2

t

z

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Common Exam - 2006

Department of PhysicsUniversity of UtahAugust 26, 2006

Examination booklets have been provided for recording your work and your solutions.

Please note that there is a separate booklet for each numbered question (i.e., usebooklet #1 for problem #1, etc.).

To receive full credit, not only should the correct solutions be given, but a sufficient

number of steps should be given so that a faculty grader can follow your reasoning.

Define all algebraic symbols that you introduce. If you are short of time it may be helpful

to give a clear outline of the steps you intended to complete to reach a solution. In some

of the questions with multiple parts you will need the answer to an earlier part in order to

work a later part. If you fail to solve the earlier part you may represent its answer with an

algebraic symbol and proceed to give an algebraic answer to the later part. This is a

closed book exam: No notes, books, or other records should be consulted. YOU MAY

ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided

equally among the ten questions of the examination.

All work done on scratch paper should be NEATLY transferred to answer booklets.

SESSION 2

COMMON EXAM DATA SHEET

e = - 1.60 × 10 C = - 4.80 × 10 esu-19 -10

c = 3.00 × 10 m/s = 3.00 × 10 cm/s8 10

h = 6.64 × 10 JAs = 6.64 × 10 ergAs = 4.14 × 10 MeVAs -34 -27 -21

S = 1.06 × 10 JAs = 1.06 × 10 ergAs = 6.59 × 10 MeVAs -34 -27 -22

k = 1.38 × 10 J/K = 1.38 × 10 erg/K -23 -16

g = 9.80 m/s = 980 cm/s2 2

G = 6.67 × 10 NAm /kg = 6.67 × 10 dyneAcm /g-11 2 2 -8 2 2

AN = 6.02 × 10 particles/gmAmole = 6.02 × 10 particles/kgAmole23 26

og (SI units) = 8.85 × 10 F/m -12

o: (SI units) = 4B × 10 H/m -7

m(electron) = 9.11 × 10 kg = 9.11 × 10 g= 5.4859 × 10 AMU = 511 keV -31 -28 -4

M(proton) 1.673 × 10 kg = 1.673 × 10 g = 1.0072766 AMU = 938.2 MeV -27 -24

M(neutron) 1.675 × 10 kg = 1.675 × 10 g = 1.0086652 AMU = 939.5 MeV -27 -24

M(muon) = 1.88 × 10 kg = 1.88 × 10 g -28 -25

1 mile = 1609 m

1 m = 3.28 ft

1 eV = 1.6 × 10 J = 1.6 × 10 ergs -19 -12

hc = 12,400 eVAD

Table of Integrals and Other Formulas

Spherical Harmonics

Conic Section

(origin at the focus; e = 1 for parabola)

Normal Distribution

Cylindrical Coordinates (orthonormal bases)

Spherical Coordinates (orthonormal bases)

Maxwell’s Equations (Rationalized MKS)

Maxwell’s Equations (Gaussian Units)

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Common exam 2006, Physics Department, University of UtahSample solutions for problem #9, Quantum Mechanics

(a) With the magnetic field in x-direction, the Hamiltonian becomes

H = −~ω0

2

(0 11 0

)so that the problem reduces to finding the eigenvalues λi and eigenvectors ψi of theσx Pauli matrix. One can obtain these eigenvalues just be looking at the matrix,however, required is here to get them formally by solving the characteristic polynomialdet [σ − λI] = 0 (I is the unity matrix). The latter reveals λ1,2 = ±1 and thereforeeigenenergies of E1,2 = ±~ω0

2and eigenvectors of

ψ1,2 =1√2

(1∓1

)if the normalization condition ψ∗ψ = 1 is taken into account.

(b) ψ(0) =

(10

)=

(1−1

)+

(11

)= 1√

2[ψ1 + ψ2]

(c) ψ(t) = e−Ht~ ψ(0) = e−

Ht~ 1√

2[ψ1 + ψ2] = 1√

2

[e−

E1t~ ψ1 + e−

E2t~ ψ2

]⇒ ψ(t) = 1

2

[e−

ω0t2

(1−1

)+ e

ω0t2

(11

)]=

(cos

(ω0t2

)i sin

(ω0t2

) )(d) p0 = p0(toff) = ψ(toff)∗ψ(0)2 = cos2

(ω0toff

2

)= 1

2[1 + cos (ω0toff)].

Interpretation: The spin precesses about the x-axis, the direction of the magnetic field~B with Larmor frequency ω, as long as the field is on.

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Problem 10 - Modern Physics:

a) [5 points] In the laboratory reference frame, what is the average 0 meson decay time? What is the average distance traveled by the 0 meson in the laboratory reference frame before it decays? The kinetic energy of the  0 is  K 0

=1.26GeV=−1m0where  is the Lorentz 

factor.  =K 0

m0

1=10 . In the laboratory reference frame, the decay time constant is affected by 

Lorentz time dilation Lab=0=10−15 s .The distance  d traveled on average by the  0

before it decays is  d=⋅c⋅Lab where  is the particle speed in unit of light speed  c . The 

Lorentz factor is  =1

1−2 and  =1− 12=1− 1

100≈0.995 which gives 

d≈3×10−7 m .

b) [6 points] From first principles, show that, in the reference frame where the 0

is at rest, the two ray photons have the same energy equal to12

m 0⋅c

2.

(1) Momentum must be conserved and, before the decay, in the rest frame, there is no momentum. So the two  rays must have momenta that are equal and opposite. (2) ray photons are massless and their energy 

is equal to their momentum multiplier by  c . This tells us the two   ray photons have the same energy. 

Finally, (3) energy must be conserved. Before the decay the energy is m0⋅c2

and each gamma ray photon 

must have an energy of 12

m 0⋅c

2. 

c) [7 points] What is the energy range within which the photons should be detected in the laboratory reference frame?

We just have obtained the energy of the  ray in the rest frame is  E=12

m0⋅c2

. With  the 

angle measured in the rest frame between the direction of the ray and that of the  0 , the component momentum of the  ray along the direction of motion of the  0 is  p x⋅c=E cos . In order to obtain the  ray energy  E ' in the laboratory frame, we use a Lorentz transform: 

E '= E Px⋅c= E E cos . It is clear that  E ' will be in the range from 12m 0

1− to 12m 0

1 . With  =10 and  =0.995 we get 

3.5MeV≤E≤1.4GeV . 

d) [7 points] Consider the ray photons that are emitted in the reference frame

where the 0 is at rest along a direction making an angle of2

with the

direction of motion of the 0 meson. In the laboratory frame, what is the angle between the direction of these photons and that of the 0 ?The transverse component of the momentum is unaffected by the Lorentz transform and 

P y '⋅c=P y⋅c=E sin . The momentum of the  measured in the laboratory was calculated 

previously  P '⋅c=E '=E 1cos . The angle measured between the direction of the 

0 and  ray in the laboratory is given by  sin ' =P ' y

P '=

sin 1 cos

. For 

=2

,  this gives sin ' =1

. With =10 , we obtain  '=sin−10.1≈5.74o

as the angle for the rays emitted at  2

in the rest frame.